How do you solve $y'' + 4x =0$ using $p=y'$ and $p(\mathrm dp/\mathrm dy) = y''\;?$

Question

How do you solve $y''+4x=0$ using $p=y'$ and $p(\mathrm dp/\mathrm dy) = y''\;?$

I am working on ODE and can't figure out how to do it without complex roots. Supposedly, I am to solve as above, where $\displaystyle x=\int\dfrac{\mathrm dy}p\,.$
However, when I set up the equation, I am stumped on how to solve. I would appreciate the help! Thank you!

Answer 1

Kai, you can solve it without any change of variables, indeed :

$\displaystyle y’=-\int4x\,\mathrm dx=-2x^2+C\;\;,$

$\displaystyle y=\int\left(-2x^2+C\right)\mathrm dx=-\dfrac23x^3+Cx+D\,.$

Answer 2

Using reduction of order should be: $p=y', p'=y''$ and then $y''+4x=0$ can be written as $p'+4x=0$ which is first order and then integrating give $p=a-2x^2$. Thus, substitution back give $y'=a-2x^2$ and then integrating give $y=Ax+B-\frac{2}{3}x^{3}$. It should be noted that this is exactly the same as the solution written by Angelo, only that I have adapted ''the substitution'' $p=y'$ here. Perhaps, it is not a good example to show the importance of order reduction using a substitution.